Let the lines 3x-4y- =0,8x-11y-33=0, and 2x-3y+ =0 be concurrent.

Question

Let the lines $3x-4y-\alpha =0,8x-11y-33=0$, and $2x-3y+\lambda =0$ be concurrent. If the image of the point
$(1,2)$ in the line $2x-3y+\lambda =0$ is $(\frac{57}{13},\frac{-40}{13})$, then $|\alpha \lambda |$ is equal to

Accepted Answer

Since $(1,2)$ reflects to $\left(\dfrac{57}{13},-\dfrac{40}{13}\right)$ across $2x-3y+\lambda=0$, the midpoint $M$ of these two points lies on the line:
$$M = \left(\frac{\frac{57}{13}+1}2,\ \frac{-\frac{40}{13}+2}2\right) = \left(\frac{35}{13},-\frac7{13}\right)$$

Substitute into $2x-3y+\lambda=0$:
$$2\left(\frac{35}{13}\right)-3\left(-\frac7{13}\right)+\lambda=0 \implies \frac{70}{13}+\frac{21}{13}+\lambda=0 \implies \lambda = -\frac{91}{13}=-7$$

Concurrency of the three lines (determinant condition) gives:
$$\begin{vmatrix}3&-4&-\alpha\8&-11&-33\2&-3&\lambda\end{vmatrix}=0$$
Expanding leads to $-\lambda+2\alpha-33=0$. With $\lambda=-7$:
$$7+2\alpha-33=0 \implies 2\alpha=26 \implies \alpha=13$$

$$|\alpha\lambda| = |13\times(-7)| = 91$$

Solutions · 1